How to get the pivot center location
Yes, you can get it without moving the cursor.

    import bpy

o = bpy.context.object
vcos = [ o.matrix_world * v.co for v in o.data.vertices ]
findCenter = lambda l: ( max(l) + min(l) ) / 2

x,y,z  = [ [ v[i] for v in vcos ] for i in range(3) ]
center = [ findCenter(axis) for axis in [x,y,z] ]

print( center )


(That was not made by me)

Will print the location of the active object.

Note, in the image it shows 
[5.960464477539063e-08, -1.1920928955078125e-07, 3.3469696044921875] instead of [0, 0, 3.3469696044921875]
 but it's a super tiny difference of at most 0.00000011920928955078125